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CHAPTER 14
STATISTICS AND PROBABILITY
(A) Main Concepts and Results
Statistics
Meaning of ‘statistics’, Primary and secondary data, Raw/ungrouped data, Range of
data, Grouped data-class intervals, Class marks, Presentation of data - frequency
distribution table, Discrete frequency distribution and continuous frequency distribution.
•
•
Graphical representation of data :
(i)
Bar graphs
(ii)
Histograms of uniform width and of varying widths
(iii)
Frequency polygons
Measures of Central tendency
(a)
Mean
(i)
Mean of raw data
n
Mean = x =
x1 + x2 + ... + xn
=
n
∑ xi
i= 1
n
where x1 , x2, ..., xn are n observations.
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EXEMPLAR PROBLEMS
(ii)
Mean of ungrouped data
x=
∑ fi xi
∑ fi
where fi’s are frequencies of xi’s.
(b) Median
A median is the value of the observation which divides the data into two equal parts,
when the data is arranged in ascending (or descending) order.
Calculation of Median
When the ungrouped data is arranged in ascending (or descending) order, the median
of data is calculated as follows :
(i)
When the number of observations (n) is odd, the median is the value of the
 n + 1


 2 
(ii)
th
observation.
When the number of observations (n) is even, the median is the average or
th
n
n 
mean of the   and  + 1
2 
 2
th
observations.
(c) Mode
The observation that occurs most frequently, i.e., the observation with maximum
frequency is called mode. Mode of ungrouped data can be determined by observation/
inspection.
Probability
•
•
•
•
Random experiment or simply an experiment
Outcomes of an experiment
Meaning of a trial of an experiment
The experimental (or empirical) probability of an event E (denoted by P(E))
is given by
P(E) =
•
Number of trials in which the event has happened
Total number of trials
The probability of an event E can be any number from 0 to 1. It can also be
0 or 1 in some special cases.
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(B) Multiple Choice Questions
Write the correct answer in each of the following :
Sample Question 1: The marks obtained by 17 students in a mathematics test (out of
100) are given below :
91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49.
The range of the data is :
(A) 46
(B) 54
(C) 90
Solution : Answer (B)
Sample Question 2: The class-mark of the class 130-150 is :
(D)
100
(A) 130
(B) 135
(C) 140
(D) 145
Solution : Answer (C)
Sample Question 3 : A die is thrown 1000 times and the outcomes were recorded as
follows :
Outcome
Frequency
1
2
3
4
5
6
180
150
160
170
150
190
If the die is thrown once more, then the probability that it shows 5 is :
(A)
9
50
(B)
3
20
(C)
4
25
(D)
7
25
(D)
120
Solution : Answer (B)
EXERCISE 14.1
Write the correct answer in each of the following :
1. The class mark of the class 90-120 is :
(A) 90
(B) 105
(C) 115
2. The range of the data :
25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is
(A) 10
(B) 15
(C) 18
(D) 26
3. In a frequency distribution, the mid value of a class is 10 and the width of the class
is 6. The lower limit of the class is :
(A) 6
(B) 7
(C) 8
(D) 12
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4. The width of each of five continuous classes in a frequency distribution is 5 and the
lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
(A) 15
(B) 25
(C) 35
(D) 40
5. Let m be the mid-point and l be the upper class limit of a class in a continuous
frequency distribution. The lower class limit of the class is :
(A) 2m + l
(B) 2m – l
(C) m – l
(D) m – 2l
6. The class marks of a frequency distribution are given as follows :
15, 20, 25, ...
The class corresponding to the class mark 20 is :
(A) 12.5 – 17.5 (B) 17.5 – 22.5 (C) 18.5 – 21.5
(D) 19.5 – 20.5
7. In the class intervals 10-20, 20-30, the number 20 is included in :
(A) 10-20
(B) 20-30
(C) both the intervals
(D) none of these intervals
8. A grouped frequency table with class intervals of equal sizes using 250-270
(270 not included in this interval) as one of the class interval is constructed for the
following data :
268, 220, 368, 258, 242, 310, 272, 342,
310, 290, 300, 320, 319, 304, 402, 318,
406, 292, 354, 278, 210, 240, 330, 316,
406, 215, 258, 236.
The frequency of the class 310-330 is:
(A) 4
(B) 5
(C) 6
(D) 7
9. A grouped frequency distribution table with classes of equal sizes using 63-72
(72 included) as one of the class is constructed for the following data :
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,
102, 110, 88, 74, 112, 14, 34, 44.
The number of classes in the distribution will be :
(A) 9
(B) 10
(C) 11
(D)
12
10. To draw a histogram to represent the following frequency distribution :
Class interval
Frequency
5-10
10-15
15-25
25-45
45-75
6
12
10
8
15
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the adjusted frequency for the class 25-45 is :
(A) 6
(B) 5
(C) 3
(D) 2
11. The mean of five numbers is 30. If one number is excluded, their mean becomes
28. The excluded number is :
(A) 28
(B) 30
(C) 35
12. If the mean of the observations :
x, x + 3, x + 5, x + 7, x + 10
is 9, the mean of the last three observations is
(A)
10
1
3
(B)
10
2
3
(C)
11
1
3
(D)
38
(D)
11
2
3
n
13. If x represents the mean of n observations x1, x2, ..., xn, then value of
( xi − x ) is:
∑
i =1
(A) –1
(B) 0
(C) 1
(D) n – 1
14. If each observation of the data is increased by 5, then their mean
(A) remains the same
(B) becomes 5 times the original mean
(C)
is decreased by 5
(D)
is increased by 5
15. Let x be the mean of x1 , x2 , ... , xn and y the mean of y1, y2, ... , yn . If z is the
mean of x1, x2, ... , xn, y1, y2, ... , yn, then z is equal to
(A)
x+y
(B)
x+y
2
(C)
x+y
n
(D)
x+y
2n
16. If x is the mean of x1, x2, ... , xn, then for a ≠ 0, the mean of ax1, ax2, ..., ax n, x1 ,
a
x2
x
, ... , n is
a
a
(A)
1

a +  x

a
(B)
1 x

a + 

a 2
(C)
1 x

a + 

a n
(D)

1
a +  x

a
2n
17. If x1 , x2 , x3 , ... , xn are the means of n groups with n 1, n 2, ... , n n number of
observations respectively, then the mean x of all the groups taken together is
given by :
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EXEMPLAR PROBLEMS
n
n
(A)
ni xi
∑
i =1
n
∑ ni xi
(B)
i=1
n
2
n
∑ ni xi
(C)
i= 1
n
∑ ni
∑ ni xi
(D)
i= 1
2n
i= 1
18. The mean of 100 observations is 50. If one of the observations which was 50 is
replaced by 150, the resulting mean will be :
(A) 50.5
(B) 51
(C) 51.5
(D) 52
19. There are 50 numbers. Each number is subtracted from 53 and the mean of the
numbers so obtained is found to be –3.5. The mean of the given numbers is :
(A) 46.5
(B) 49.5
(C) 53.5
(D) 56.5
20. The mean of 25 observations is 36. Out of these observations if the mean of first
13 observations is 32 and that of the last 13 observations is 40, the 13th observation
is :
(A) 23
(B) 36
(C) 38
(D) 40
21. The median of the data
78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
(A) 45
(B) 49.5
(C) 54
(D) 56
22. For drawing a frequency polygon of a continous frequency distribution, we plot the
points whose ordinates are the frequencies of the respective classes and abcissae
are respectively :
(A) upper limits of the classes
(C) class marks of the classes
23. Median of the following numbers :
4, 4, 5, 7, 6, 7, 7, 12, 3 is
(A) 4
(B) 5
(B)
(D)
lower limits of the classes
upper limits of perceeding classes
(C)
6
(D)
7
24. Mode of the data
15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is
(A) 14
(B) 15
(C) 16
(D) 17
25. In a sample study of 642 people, it was found that 514 people have a high school
certificate. If a person is selected at random, the probability that the person has a
high school certificate is :
(A) 0.5
(B) 0.6
(C) 0.7
(D) 0.8
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26. In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat
potato chips. If a child is selected at random, the probability that he/she does not
like to eat potato chips is :
(D) 0.80
(C) 0.75
(B) 0.50
(A) 0.25
27. In a medical examination of students of a class, the following blood groups are
recorded:
Blood group
A
AB
B
O
Number of students
10
13
12
5
A student is selected at random from the class. The probability that he/she has
blood group B, is :
1
3
13
1
(D)
(C)
(B)
8
10
40
4
28. Two coins are tossed 1000 times and the outcomes are recorded as below :
(A)
Number of heads
Frequency
2
1
0
200
550
250
Based on this information, the probability for at most one head is
3
4
1
1
(D)
(C)
(B)
4
5
4
5
29. 80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in
the form of a frequency table given below :
(A)
Life time (in hours)
300
500
700
900
1100
Frequency
10
12
23
25
10
One bulb is selected at random from the lot. The probability that its life is 1150
hours, is
(A)
1
80
(B)
7
16
(C)
0
(D)
1
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30. Refer to Q.29 above :
The probability that bulbs selected randomly from the lot has life less than 900
hours is :
(A)
11
40
(B)
5
16
7
16
(C)
(D)
9
16
(C) Short Answer Questions with Reasoning
Sample Question 1 : The mean of the data :
2, 8, 6, 5, 4, 5, 6, 3, 6, 4, 9, 1, 5, 6, 5
is given to be 5. Based on this information, is it correct to say that the mean of the data:
10, 12, 10, 2, 18, 8, 12, 6, 12, 10, 8, 10, 12, 16, 4
is 10? Give reason.
Solution : It is correct. Since the 2nd data is obtained by multiplying each observation
of 1st data by 2, therefore, the mean will be 2 times the mean of the 1st data.
Sample Question 2 : In a histogram, the areas of the rectangles are proportional to
the frequencies. Can we say that the lengths of the rectangles are also proportional to
the frequencies?
Solution: No. It is true only when the class sizes are the same.
Sample Quetion 3 : Consider the data : 2, 3, 9, 16, 9, 3, 9. Since 16 is the highest value
in the observations, is it correct to say that it is the mode of the data? Give reason.
Solution : 16 is not the mode of the data. The mode of a given data is the observation
with highest frequency and not the observation with highest value.
EXERCISE 14.2
1. The frequency distribution :
Marks
Number of Students
0-20
20-40
40-60
60-100
10
15
20
25
has been represented graphically as follows :
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Fig. 14.1
Do you think this representation is correct? Why?
2. In a diagnostic test in mathematics given to students, the following marks (out of
100) are recorded :
46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44
Which ‘average’ will be a good representative of the above data and why?
3. A child says that the median of 3, 14, 18, 20, 5 is 18. What doesn’t the child
understand about finding the median?
4. A football player scored the following number of goals in the 10 matches :
1, 3, 2, 5, 8, 6, 1, 4, 7, 9
Since the number of matches is 10 (an even number), therefore, the median
=
5 th observation + 6 th observation
2
8+6
=7
2
Is it the correct answer and why?
5. Is it correct to say that in a histogram, the area of each rectangle is proportional to
the class size of the corresponding class interval? If not, correct the statement.
=
6. The class marks of a continuous distribution are :
1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and 1.64
Is it correct to say that the last interval will be 1.55 - 1.73? Justify your answer.
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EXEMPLAR PROBLEMS
7. 30 children were asked about the number of hours they watched TV programmes
last week. The results are recorded as under :
Number of hours
Frequency
0-5
5-10
10-15
15-20
8
16
4
2
Can we say that the number of children who watched TV for 10 or more hours a
week is 22? Justify your answer.
8. Can the experimental probability of an event be a negative number? If not, why?
9. Can the experimental probability of an event be greater than 1? Justify your anwer.
10. As the number of tosses of a coin increases, the ratio of the number of heads to the
total number of tosses will be
1
. Is it correct? If not, write the correct one.
2
(D) Short Answer Questions
Sample Question 1 : Heights (in cm) of 30 girls of Class IX are given below:
140, 140, 160, 139, 153, 153, 146, 150, 148, 150, 152,
146, 154, 150, 160, 148, 150, 148, 140, 148, 153, 138,
152, 150, 148, 138, 152, 140, 146, 148.
Prepare a frequency distribution table for this data.
Solution : Frequency distribution of heights of 30 girls
Height
(in cm)
Tally Marks
138
139
140
146
148
150
152
153
154
160
||
|
||||
|||
|||| |
||||
|||
|||
|
||
Frequency
2
1
4
3
6
5
3
3
1
2
Total 30
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Sample Question 2 : The following observations are arranged in ascending order :
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93
If the median is 65, find the value of x.
Solution : Number of observations (n) = 10, which is even. Therefore, median is the
th
th
 n
n 
mean of   and  + 1 observation, i.e., 5th and 6th observation.
 2
2 
Here,
5th observation = x
6th observation = x + 2
Median =
x + ( x + 2)
= x +1
2
Now,
x + 1 = 65 (Given)
Therefore,
x = 64
Thus, the value of x is 64.
Sample Question 3 : Here is an extract from a mortality table.
Age (in years)
60
61
62
63
64
65
(i)
Number of persons surviving out
of a sample of one million
16090
11490
8012
5448
3607
2320
Based on this information, what is the probability of a person ‘aged 60’ of
dying within a year?
(ii) What is the probability that a person ‘aged 61’ will live for 4 years?
Solution :
(i) We see that 16090 persons aged 60, (16090-11490), i.e., 4600 died before
reaching their 61st birthday.
Therefore, P(a person aged 60 die within a year) =
4600
460
=
16090
1609
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EXEMPLAR PROBLEMS
(ii)
Number of persons aged 61 years = 11490
Number of persons surviving for 4 years = 2320
P(a person aged 61 will live for 4 years) =
2320
232
=
11490 1149
EXERCISE 14.3
1. The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B,
A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
2. The value of π upto 35 decimal places is given below:
3. 14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.
3. The scores (out of 100) obtained by 33 students in a mathematics test are as
follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84 000
66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.
4. Prepare a continuous grouped frequency distribution from the following data:
Mid-point
Frequency
5
4
15
8
25
13
35
12
45
6
Also find the size of class intervals.
5. Convert the given frequency distribution into a continuous grouped frequency
distribution:
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Class interval
Frequency
150-153
154-157
158-161
7
7
15
162-165
166-169
170-173
10
5
6
In which intervals would 153.5 and 157.5 be included?
6. The expenditure of a family on different heads in a month is given below:
Head
Food
Expenditure
(in Rs)
4000
Education Clothing House Rent Others
2500
1000
3500
2500
Savings
1500
Draw a bar graph to represent the data above.
7. Expenditure on Education of a country during a five year period (2002-2006), in
crores of rupees, is given below:
Elementary education
Secondary Education
University Education
Teacher’s Training
Social Education
Other Educational Programmes
Cultural programmes
Technical Education
240
120
190
20
10
115
25
125
Represent the information above by a bar graph.
8. The following table gives the frequencies of most commonly used letters a, e, i, o,
r, t, u from a page of a book :
Letters
a
e
i
o
r
t
u
Frequency
75
125
80
70
80
95
75
Represent the information above by a bar graph.
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EXEMPLAR PROBLEMS
9. If the mean of the following data is 20.2, find the value of p:
x
10
15
20
25
30
f
6
8
p
10
6
10. Obtain the mean of the following distribution:
Frequency
4
8
14
11
3
Variable
4
6
8
10
12
11. A class consists of 50 students out of which 30 are girls. The mean of marks
scored by girls in a test is 73 (out of 100) and that of boys is 71. Determine the
mean score of the whole class.
12. Mean of 50 observations was found to be 80.4. But later on, it was discovered that
96 was misread as 69 at one place. Find the correct mean.
13. Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in an
ascending order. The median of the data is 24. Find the value of x.
14. The points scored by a basket ball team in a series of matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28
Find the median and mode for the data.
15. In Fig. 14.2, there is a histogram depicting daily wages of workers in a factory.
Construct the frequency distribution table.
Fig. 14.2
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16. A company selected 4000 households at random and surveyed them to find out a
relationship between income level and the number of television sets in a home. The
information so obtained is listed in the following table:
Monthly income
(in Rs)
Number of Televisions/household
0
1
2
Above 2
< 10000
10000 - 14999
15000 - 19999
20000 - 24999
20
10
0
0
80
240
380
520
10
60
120
370
0
0
30
80
25000 and above
0
1100
760
220
Find the probability:
(i) of a household earning Rs 10000 – Rs 14999 per year and having exactly
one television.
(ii) of a household earning Rs 25000 and more per year and owning 2 televisions.
(iii) of a household not having any television.
17. Two dice are thrown simultaneously 500 times. Each time the sum of two numbers
appearing on their tops is noted and recorded as given in the following table:
Sum
Frequency
2
14
3
30
4
42
5
55
6
72
7
75
8
70
9
53
10
46
11
28
12
15
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EXEMPLAR PROBLEMS
If the dice are thrown once more, what is the probability of getting a sum
(i)
3?
(ii)
more than 10?
(iii) less than or equal to 5?
(iv) between 8 and 12?
18. Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons
were examined for defective bulbs and the results are given in the following table:
Number of defective bulbs
Frequency
0
1
2
3
4
400 180 48 41 18
5
6 more than 6
8
3
2
One carton was selected at random. What is the probability that it has
(i)
no defective bulb?
(ii)
defective bulbs from 2 to 6?
(iii) defective bulbs less than 4?
19. Over the past 200 working days, the number of defective parts produced by a
machine is given in the following table:
Number of
0
1
2
3
4
5
6
7
8
9 10 11 12 13
defective parts
Days
50 32 22 18 12 12 10 10 10 8
6
6
2
2
Determine the probability that tomorrow’s output will have
(i)
no defective part
(ii)
atleast one defective part
(iii) not more than 5 defective parts
(iv) more than 13 defective parts
20. A recent survey found that the ages of workers in a factory is distributed as follows:
Age (in years)
Number of workers
20 - 29 30 - 39 40 - 49 50 - 59
38
27
86
46
60 and above
3
If a person is selected at random, find the probability that the person is:
(i)
40 years or more
(ii)
under 40 years
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(iii)
(iv)
145
having age from 30 to 39 years
under 60 but over 39 years
(E) Long Answer Questions
Sample Question 1: Following is the frequency distribution of total marks obtained
by the students of different sections of Class VIII.
Marks
100 - 150 150 - 200 200 - 300 300 - 500 500 - 800
Number of students
60
100
100
80
180
Draw a histogram for the distribution above.
Solution: In the given frequency distribution, the class intervals are not of equal width.
Therefore, we would make modifications in the lengths of the rectangles in the histogram
so that the areas of rectangles are proportional to the frequencies. Thus, we have:
Marks
Frequency
Width of the class
Length of the rectangle
100 - 150
60
50
50
× 60 = 60
50
150 - 200
100
50
50
× 100 = 100
50
200 - 300
100
100
50
×100 = 50
100
300 - 500
80
200
50
× 80 = 20
200
500 - 800
180
300
50
×180 = 30
300
Now, we draw rectangles with lengths as given in the last column. The histogram
of the data is given below :
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EXEMPLAR PROBLEMS
Fig. 14.3
Sample Question 2 : Two sections of Class IX having 30 students each appeared for
mathematics olympiad. The marks obtained by them are shown below:
46 31 74 68 42 54 14 61 83 48 37 26 8
64 57
93 72 53 59 38 16 88 75 56 46 66 45 61 54 27
27 44 63 58 43 81 64 67 36 49 50 76 38 47 55
77 62 53 40 71 60 58 45 42 34 46 40 59 42 29
Construct a group frequency distribution of the data above using the classes 0-9, 10-19
etc., and hence find the number of students who secured more than 49 marks.
Solution :
Class
0-9
10-19
20-29
30-39
Tally Marks
|
||
||||
|||| |
40-49
50-59
60-69
70-79
80-89
||||
||||
||||
||||
|||
90-99
|
Total
|||| ||||
|||| ||
||||
|
Frequency
1
2
4
6
15
12
10
6
3
1
60
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From the table above, we find that the number of students who secure more than 49
marks is (12 + 10 + 6 + 3 + 1), i.e., 32.
EXERCISE 14.4
1. The following are the marks (out of 100) of 60 students in mathematics.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75,
31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62,
52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30.
Construct a grouped frequency distribution table with width 10 of each class starting
from 0 - 9.
2. Refer to Q1 above. Construct a grouped frequency distribution table with width 10
of each class, in such a way that one of the classes is 10 - 20 (20 not included).
3. Draw a histogram of the following distribution :
Heights (in cm)
Number of students
150 - 153
153 - 156
7
8
156 - 159
159 - 162
162 - 165
165 - 168
14
10
6
5
4. Draw a histogram to represent the following grouped frequency distribution :
Ages (in years)
Number of teachers
20 - 24
25 - 29
30 - 34
35 - 39
10
28
32
48
40 - 44
45 - 49
50 - 54
50
35
12
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EXEMPLAR PROBLEMS
5. The lengths of 62 leaves of a plant are measured in millimetres and the data is
represented in the following table :
Length (in mm)
Number of leaves
118 - 126
127 - 135
136 - 144
145 - 153
154 - 162
163 - 171
172 - 180
8
10
12
17
7
5
3
Draw a histogram to represent the data above.
6. The marks obtained (out of 100) by a class of 80 students are given below :
Marks
Number of students
10 - 20
6
20 - 30
30 - 50
50 - 70
70 - 100
17
15
16
26
Construct a histogram to represent the data above.
7. Following table shows a frequency distribution for the speed of cars passing through
at a particular spot on a high way :
Class interval (km/h)
Frequency
30 - 40
40 - 50
3
6
50 - 60
60 - 70
70 - 80
80 - 90
90 - 100
25
65
50
28
14
Draw a histogram and frequency polygon representing the data above.
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149
8. Refer to Q. 7 :
Draw the frequency polygon representing the above data without drawing the
histogram.
9. Following table gives the distribution of students of sections A and B of a class
according to the marks obtained by them.
Section A
Marks
0 - 15
15 - 30
30 - 45
45 - 60
60 - 75
75 - 90
Frequency
5
12
28
30
35
13
Section B
Marks
0 - 15
15 - 30
30 - 45
45 - 60
60 - 75
75 - 90
Frequency
3
16
25
27
40
10
Represent the marks of the students of both the sections on the same graph by two
frequency polygons.What do you observe?
10. The mean of the following distribution is 50.
x
f
10
30
50
70
90
17
5a + 3
32
7a – 11
19
Find the value of a and hence the frequencies of 30 and 70.
11. The mean marks (out of 100) of boys and girls in an examination are 70 and 73,
respectively. If the mean marks of all the students in that examination is 71, find
the ratio of the number of boys to the number of girls.
12. A total of 25 patients admitted to a hospital are tested for levels of blood sugar,
(mg/dl) and the results obtained were as follows :
87
71
83
67
85
77
69
76
65
85
85
54
70
68
80
73
78
68
85
73
81
78
81
77
75
Find mean, median and mode (mg/dl) of the above data.
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